Advanced Level Organic Chemistry: Free radical reactions of benzene & methylbenzene

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Part 7.4 Free radical addition reactions of benzene and methylbenzene with hydrogen and chlorination substitution of the methyl group of methylbenzene

Sub-index for this page

7.4.1 Hydrogenation of benzene and methylbenzene

7.4.2 Free radical addition of chlorine to benzene

7.4.3 Free radical substitution by chlorine of the methyl group in methylbenzene

7.4.1  Hydrogenation of benzene and methylbenzene

If benzene or methylbenzene is mixed with hydrogen and passed over a heated nickel catalyst addition of three moles of hydrogen per arene molecule takes place.

The higher temperature and Ni catalyst are needed because of the stability of the benzene ring in arenes.

The hydrogen molecules are adsorbed on the Ni surface and split into atoms.

These atoms are effectively very reactive hydrogen radicals (H•) which can break open the pi bond system of the arene and form new C-H bonds until the totally saturated cycloalkane is formed.


(1)   +  3H2 ===>  alkanes structure and naming (c) doc b    or    skeletal formula of benzene C6H6 + 3H2 ===> alkanes structure and naming (c) doc b

hydrogenation of the benzene to cyclohexane.

ΔHhydrogenation(benzene) = -208 kJ mol-1


(2) (c) doc b  +  3H2  ===> alkanes structure and naming (c) doc b   or    skeletal formula of methylbenzene C7H8 +  3H2  ===> 

hydrogenation of methylbenzene to methylcyclohexane

ΔHhydrogenation(methylbenzene) = -205 kJ mol-1

Note the similarity of the enthalpies of hydrogenation.

diagram free radical catalyttic hydrogenation of benzene or methylbenzene nickel surface catalyst

The mechanism is very complicated, but the very simplified diagram above gives you some about it.

Hydrogen free radicals (hydrogen atoms with their electron) are formed on the nickel surface, and then attack the pi electron cloud of the 'benzene' ring forming new C-H bonds.

A series of free radical chain reactions ensue until all the pi electrons have been used in forming C-H bonds and the fully saturated cycloalkane is formed.

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7.4.2 Free radical addition of chlorine to benzene

If chlorine is bubbled into boiling liquid benzene as ultraviolet light is shone on the mixture you get the addition of three chlorine molecules per benzene molecule.

Unlike alkenes, addition will not take place without the input of high energy uv light to trigger a free radical reaction.

You do get addition under these particular reaction conditions.

The reaction yield yields 1,2,3,4,5,6-hexachlorocyclohexane.

C6H6  +  3Cl2  ===>  C6H6Cl6

diagram uv catalysed free radical addition of chlorine to benzene giving 1,2,3,4,5,6-hexachlorocyclohexane


Cl2  == uv  ==> 2Cl•

Propagation steps

C6H6  +  Cl•  ===> C6H6Cl•

etc. but not sure of mechanism? (can't find it on the internet)

until the final step is maybe?

C6H6Cl5•  +  Cl•  ===> C6H6Cl6 

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7.4.3 Free radical substitution by chlorine of the methyl group in methylbenzene

When methylbenzene is reacted with chlorine and the mixture irradiated with uv light, you do NOT get addition like you do with benzene!

Instead, you get a free radical substitution reaction involving the methyl group - a side-chain substitution.

The initial product is (chloromethyl)benzene (benzyl chloride).

(i) methylbenzene  +  chlorine  ===> (chloromethyl)benzene  +  hydrogen chloride

C6H5CH3  +  Cl2  == uv ==> C6H5CH2Cl  +  HCl

(c) doc b   +  Cl2  ===>  (c) doc b  +  HCl

(chloromethyl)benzene is not an aryl halide, but a primary aliphatic halogenoalkane.


With excess chlorine, the free radical chain reaction can continue to form two other substitution products.


(ii) (chloromethyl)benzene  +  chlorine  ===> (dichloromethyl)benzene  +  hydrogen chloride

C6H5CH2Cl  +  Cl2  ===>  C6H5CHCl2  +  HCl


(iii) (dichloromethyl)benzene  +  chlorine  ===> (trichloromethyl)benzene  +  hydrogen chloride

C6H5CHCl2  +  Cl2  ===>  C6H5CCl3  +  HCl


Again, these are all aliphatic halogenoalkanes, NOT aryl halides.


A similar reaction will happen with bromine to give three possible bromo-substituted products and via  similar free radical chain mechanism outlined below..

The free radical mechanism of the side-chain chlorination of methylbenzene

(• = unpaired electron on a free radical)

Initiation step

Cl2  == uv  ==> 2Cl•

Chain propagation steps

C6H5CH3  +  Cl•  ===>  C6H5CH2•  +  HCl

C6H5CH2•  +  Cl2  ===>  C6H5CH2Cl  +  Cl•   (1st product, monochloro...)

Termination steps

2Cl•  ===> Cl2

C6H4CH2•  +  Cl•  ===>  C6H5CH2Cl

2C6H4CH2•  ===>  C6H5CH2C6H5

Further propagation steps to form the 2nd and 3rd substitution products.

C6H5CH2Cl  +  Cl•  ===>  C6H5CHCl•  +  HCl

C6H5CHCl•  +  Cl2  ===>  C6H5CHCl2  +  Cl•   (2nd product, dichloro...)

C6H5CHCl2  +  Cl•  ===>  C6H5CCl2•  +  HCl

C6H5CCl2•  +  Cl2  ===>  C6H5CCl3  +  Cl•   (3rd product, trichloro...)


As far as I know its the same set of reactions for bromine.


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